Problem: You have found the following ages (in years) of all 4 turtles at your local zoo: $ 90,\enspace 13,\enspace 57,\enspace 4$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 4 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{90 + 13 + 57 + 4}{{4}} = {41\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $90$ years $49$ years $2401$ years $^2$ $13$ years $-28$ years $784$ years $^2$ $57$ years $16$ years $256$ years $^2$ $4$ years $-37$ years $1369$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{2401} + {784} + {256} + {1369}} {{4}} $ $ {\sigma^2} = \dfrac{{4810}}{{4}} = {1202.5\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{1202.5\text{ years}^2}} = {34.7\text{ years}} $ The average turtle at the zoo is 41 years old. There is a standard deviation of 34.7 years.